Using Gauss’ law, find the approximate magnitude of the electric field at the surface of a cube that has 0.10-m sides and a uniform volume charge density ρ= 2.0 ×10−9C/m3.

How to solve it?

The approximate magnitude of the electric field at the surface of the cube is 1.8 N/C.

Explanation:

The electric field, E, at the surface of the cube can be determined by:

                  E =

Where k is a constant (8.99 x 109 ), q is the charge and r the distance from the charge.

But the volume charge density, ρ, is given as:

                 ρ =

where: q is the electric charge and V the volume.

From the question, sides of cube = 0.10 m and uniform volume charge density, ρ = 2.0 × .

Thus:

volume of cube = length ×length × length

                          = 0.10 ×0.10 ×0.10

                          = 1.0 ×

So that:

            q = ρ × V

               = 2.0 × × 1.0 ×

               = 2.0 ×

            q = 2.0 × C

The electric field, E, at the surface of the cube can be determined by:

                  E =

k = 8.99 x 109 ), q = 2.0 × C, r = 0.1 m, so that:

                E =

                   = 1.798 N/C

The approximate magnitude of the electric field at the surface of the cube is 1.8 N/C.

double the distance = divide strength by 4